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Suppose a manifold is covered as the union of $|A|$-many $n$-spheres, i.e. $M=\bigcup_{\alpha\in A}S^n$. We are allowed to move the spheres around so long as they cover $M$; that is, we can make the intersection as small as possible (e.g. a point). I would like to compute the homology groups of $M$; however, we have no information regarding the intersections of of the spheres (other than the fact that they cannot be disjoint).

Is there a way to "arrange" the spheres in such a way that they still cover $M$ but have homology that is computable? Could I contract the intersection to a point and then compute the homology of a wedge sum $H_k\left(\bigvee_{\alpha\in A} S^n\right)=\bigoplus_{\alpha\in A} H_k(S^n)$?

Note, $A$ can be made to be countably infinite or uncountably infinite.

Thanks in advance! Any help would be much appreciated.

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    $\begingroup$ I'm not sure I understand but when each $U_\alpha$ is homeomorphic to an open ball then the connected sum $U\alpha \#U_\beta$ is always a ball. So $\#U_\alpha$ would be a ball, right? $\endgroup$
    – freakish
    Aug 22 '18 at 17:21
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    $\begingroup$ At some point, you're going to have to attach a ball along something which is not a disk. $\endgroup$ Aug 22 '18 at 21:51
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    $\begingroup$ @Multivariablecalculus: try your idea on a sphere, or even better, a circle. $\endgroup$ Aug 22 '18 at 21:59
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    $\begingroup$ @Multivariablecalculus: Consider covering the circle with three overlapping coordinate charts (which are copies of open intervals.) Call them $U_1,U_2,U_3$ and assume $U_i\cap U_j$ is a small interval but the triple intersection is empty. Now you can say $U_1\cup U_2$ is homeomorphic to $U_1\#U_2$. However once you try to add $U_3$ in, it overlaps with $U_1\cup U_2$ in two different subintervals. By definition of connect sum, you can't attach to both simultaneously. $\endgroup$ Aug 23 '18 at 0:37
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    $\begingroup$ @Multivariablecalculus: yes. You can build any smooth manifold up using a handle decomposition. The wikipedia article has some info, though I don't know how easy it is to understand if you've never seen it before. $\endgroup$ Aug 23 '18 at 0:44
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For your new question, homology can be computed from the nerve of certain coverings by open balls, using the Nerve Theorem. You need all intersections of finitely many open balls to be either empty or contractible.

Edit: Here's a bit more information. You can define the nerve of any covering of a topological space by open sets. This is a simplicial complex that has a vertex for every open set in the cover and a simplex whenever a collection of finitely many open sets has a nonempty intersection. If these finite intersections are all contractible or empty, then we say the cover is a good cover. The Nerve Theorem states that the nerve of a good cover is homotopy-equivalent to the original space. Hence it will have the same homology. Admittedly, the nerve is often not any easier to compute than the original space. For my example of covering the circle by three intervals, the nerve is just a triangle, which is actually homeomorphic to the original circle.

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  • $\begingroup$ Thanks @CheerfulParsnip! Sorry to bother you, but do you mind elaborating on this answer a bit more as I only have a vague idea of what it means. $\endgroup$ Aug 23 '18 at 2:21
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    $\begingroup$ @Multivariablecalculus: Have you seen simplicial complexes? $\endgroup$ Aug 23 '18 at 2:26
  • $\begingroup$ Yes. I am just confused by the meaning of the nerve of certain coverings by open balls. @CheerfulParsnip $\endgroup$ Aug 23 '18 at 2:44
  • $\begingroup$ I have one final question. How do we do this explicitly for $\bigcup_{\alpha\in A}U_{\alpha}$ while retaining a cover of $M$? @CheerfulParsnip $\endgroup$ Aug 23 '18 at 17:49

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